The area of a trapezoid is found using the formula \( \frac{ 1 } {2} h \times (a + b ) \), where \( a \) and \( b \) are length of the parallel sides and \( h \) is the perpendicular height of the trapezoid.

Isosceles Trapezoids

An isosceles trapezoid is a trapezoid that has congruent legs.

Thus, for the isosceles trapezoid \(ABCD\) in the figure above, the following correspond:

(1) The bases are parallel.
(2) The legs are equal in length. Thus, \(\lvert\overline{BA}\rvert=\lvert\overline{CD}\rvert.\)
(3) The angles the two legs make with a base are equal. Thus, \(\angle BAD=\angle CDA\) and \(\angle ABC=\angle DCB.\)

Let points \( A^{'}\) and \(D^{'}\) be the perpendicular foots on \(\overline{BC}\) from points \(A\) and \(D,\) respectively. Then we have \(\lvert\overline{AD}\rvert=\lvert\overline{A'D'}\rvert.\)

Since the problem states that \(ABCD\) is an isosceles trapezoid, we know that \(\lvert\overline{BA'}\rvert=\lvert\overline{CD'}\rvert.\) Given that the length of \(\overline{BC}\) is twice the length of \(\overline{AD},\) we have
\[ \begin{align}
\lvert \overline{BA^{'}} \rvert &=\frac{1}{2}\cdot \lvert \overline{A^{'}D^{'}} \rvert
\end{align} \]
and since the length ratio of the hypotenuse and the base of \( \triangle ABA^{'} \) is \( 2:1 ,\) \( \angle ABA^{'} \) is \( 60^\circ . \qquad (1) \)

We know that \( \triangle ABD \) is an isosceles triangle, so \( \angle ABD=\angle ADB,\) and since \( \overline{AD} \parallel\overline{BC},\) it follows that \( \angle DBC=\angle ADB.\)

Thus,
\[ \angle ABD = \angle ADB = \angle DBC . \qquad (2) \]

From (1) and (2), we know that \( \angle ABD=\angle DBA^{'}\) and \( \angle ABA^{'} = 60^\circ.\) Hence we have